Question

The value of $\underset{n\to \infty }{lim}\cos \left(\frac{x}{2}\right)\cos \left(\frac{x}{4}\right)\cos \left(\frac{x}{8}\right)...\cos \left(\frac{x}{2^n}\right)$ is

• A. $1$
• B. $\frac{\sin x}{x}$
• C. $\frac{x}{\sin x}$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{\sin x}{x}$

$\sin x=2^1\sin \frac{x}{2}\cos \frac{x}{2}$

$\sin x=2^2\sin \frac{x}{4}\cos \frac{x}{4}\cos \frac{x}{2}$

$\sin x=2^3\sin \frac{x}{8}\cos \frac{x}{8}\cos \frac{x}{4}\cos \frac{x}{2}$

.

.

.

.

$\sin x=2^n\sin \frac{x}{2^n}{\sum }_{k=1}^n\cos \frac{x}{2^k}$

We know that

$\frac{\sin x}{x}\to 1$ as $x\to 0$

$\underset{x\to \infty }{lim}{\sum }_{k=1}^n\cos \frac{x}{2^k}=\underset{x\to \infty }{lim}\frac{\sin x}{x}{\left(\frac{\sin 2^{-n}x}{2^{-n}x}\right)}^{-1}=\frac{\sin x}{x}$